Algebric Sets

For the $n^{th}$ set we see that it contains the numbers from $n^2$ to $n^2 + n$

Therefore the sum of the numbers in the $n^{th}$ set is:

$n^2 + \left(n^2+1\right) + \left(n^2 + 2 \right) + ... + \left(n^2 + n \right)$

$= n^2 \left(n+1\right) + \frac{1}{2}n\left(n+1\right)$

$= \frac{1}{2}\left(n+1\right)\left(2n^2+n\right)=\frac{1}{2}n\left(n+1\right)\left(2n+1\right)$

Therefore, we see that $\sqrt(3)\frac{S_{50}}{\sqrt{S_{51}-S_{50}}}$

$= \sqrt(3)\frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{\sqrt{\frac{1}{2}\left(51\right)\left(52\right)\left(103\right) - \frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}}$

$= \sqrt{3}\frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{\sqrt{\frac{1}{2}\left(51\right)\left(52*103 - 50*101\right)}}$

$= \sqrt{3}\frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{\sqrt{\frac{1}{2}\left(51\right)\left(5356-5050\right)}}$

$= \sqrt{3}\frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{\sqrt{\frac{1}{2}\left(51\right)\left(306\right)}}$

$= \sqrt{3}\frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{\sqrt{\left(51\right)\left(51\right)\left(3\right)}}$

$= \frac{\frac{1}{2}\left(50\right)\left(51\right)\left(101\right)}{51}$

$= 25*101 = 2525$

First term of each set is square of the number of the set so the first term of 50th set will be 2500

And last term of the nth set is given by n(n+1) so the last term of the 50th set will be 2550

All the terms will be in $AP$

thus the sum of the terms of the 50th set will be 128775

Similarly doing in the 51th set the first term will be $51^{2}$ =2601

And the last term 2654

And all these are in AP THUS THE SUM OF ALL THE terms will be 136578

So we have got S(50) and S(51) on putting in the question we get s51-s50=7803

$\sqrt{7803}$ =51 $\sqrt{3}$ ----------(S51-S50)

So final answer is

$\frac{128775}{51}$ $\sqrt{3}$ * $\sqrt{3}$ = $\boxed{2525}$

This is our answer

Observe sets.....each set has one element more than its position and highest value is position * no of elements.... So s50 has 51 elements starting from 2500 to 2550 and s51 has elements starting from 2601 to 2652(51*52)....find both sums using arithmetic progression and substitute. And is 2525.